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**Isomorphism** $\displaystyle d^2 = ( x - 1 )^2 + ( 3x + 3 )^2 $

$\displaystyle = 10x^2 + 16x + 10 $

$\displaystyle = (\sqrt{10}x)^2 + 2\left(\frac{8}{\sqrt{10}}\right)(\sqrt{10}x) + \left(\frac{8}{\sqrt{10}}\right)^2 + \left(10 - \frac{64}{10}\right) $

$\displaystyle = \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 + \frac{36}{10}$

$\displaystyle \boxed{d^2 \geq \frac{36}{10}}$ since $\displaystyle \left(\sqrt{10}x + \frac{8}{\sqrt{10}}\right)^2 \geq 0$

$\displaystyle d^2 \geq \frac{36}{10} \implies d \geq \frac{6}{\sqrt{10}}$

Now show that $\displaystyle \frac{6}{\sqrt{10}} = \frac{3\sqrt{10}}{5}$ and you are done.